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## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.8

**9th Maths Exercise 3.8 Question 1.**

Factorise each of the following polynomials using synthetic division:

(i) x^{3} – 3x^{2} – 10x + 24

(ii) 2x^{3} – 3x^{2} – 3x + 2

(iii) -7x + 3 + 4x^{3}

(iv) x^{3} + x^{2} – 14x – 24

(v) x^{3} – 7x + 6

(vi) x^{3} – 10x^{2} – x + 10

Solution:

(i) x^{3} – 3x^{2} – 10x + 24

Let p(x) = x^{3} – 3x^{2} – 10x + 24

Sum of all the co-efficients = 1 – 3 – 10 + 24 = 25 – 13 = 12 ≠ 0

Hence (x – 1) is not a factor.

Sum of co-efficient of even powers with constant = -3 + 24 = 21

Sum of co-efficients of odd powers = 1 – 10 = – 9

21 ≠ -9

Hence (x + 1) is not a factor.

p (2) = 2^{3} – 3 (2^{2}) – 10 × 2 + 24 = 8 – 12 – 20 + 24

= 32 – 32 = 0 ∴ (x – 2) is a factor.

Now we use synthetic division to find other factor

Thqs (x – 2) (x + 3) (x – 4) are the factors.

∴ x^{3} – 3x^{2} – 10x + 24 = (x – 2) (x + 3) (x – 4)

(ii) 2x^{2} – 3x^{2} – 3x + 2

Let p (x) = 2x^{3} – 3x^{2} – 3x + 2

Sum of all the co-efficients are

2 – 3 – 3 + 2 = 4 – 6 = -2 ≠ 0

∴ (x – 1) is not a factor

Sum of co-efficients of even powers of x with constant = -3 + 2 = – 1

Sum of co-efficients of odd powers of x = 2- 3= -1

(-1) = (-1)

∴ (x + 1) is a factor

Let us find the other factors using synthetic division

Quotient is 2x^{2} – 5x + 2 = 2x – 4x – x + 2 = 2x (x – 2) – 1 (x – 2)

= (x – 2) (2x – 1)

∴ 2x^{3} – 3x^{2} – 3x + 2 = (x + 1) (x – 2) (2x – 1)

(iii) -7x + 3 + 4x^{3}

Let p(x) = 4x^{3} + 0x^{2} – 7x + 3

Sum of the co-efficients are = 4 + 0 – 7 + 3

= 7 – 7 = 0

∴ (x- 1) is a factor

Sum of co-efficients of even powers of x with constant = 0 + 3 = 3

Sum of co-efficients of odd powers of x with constant = 4 – 7 = -3

-3 ≠ -3

∴ (x + 1) is not a factor

Using synthetic division, let us find the other factors.

Quotient is 4x^{2} + 4x – 3

= 4x^{2} + 6x – 2x – 3

= 2x (2x + 3) – 1 (2x + 3)

= (2x + 3) (2x – 1)

∴ The factors are (x – 1), (2x + 3) and (2x – 1)

∴ -7x + 3 + 4x^{3} = (x + 1) (2x + 3) (2x – 1)

(iv) x^{3} + x^{2} – 14x – 24

Let p (x) = x^{3} + x^{2} – 14x – 24

Sum of the co-efficients are = 1 + 1 – 14 – 24 = -36 ≠ 0

∴ (x – 1) is not a factor

Sum of co-efficients of even powers of x with constant = 1 – 24 = -23

Sum of co-efficients of odd powers of x = 1 – 14 = -3

-23 ≠ -13

∴ (x + 1) is also not a factor

p(2) = 2^{3} + 2^{2} – 14 (2) – 24 = 8 + 4 – 28 – 24

= 12 – 52 ≠ 0, (x – 2) is a not a factor

p (-2) = (-2)^{3} + (-2)^{2} – 14 (-2) – 24

= -8 + 4 + 28 – 24 = 32 – 32 = 0

∴ (x + 2) is a factor

To find the other factors let us use synthetic division.

∴ The factors are (x + 2), (x + 3), (x + 4)

∴ x^{3} + x^{2} – 14x – 24 = (x + 2) (x + 3) (x – 4)

(v) x^{3} – 7x + 6

Let p (x) = x^{3} + 0x^{2} – 7x + 6

Sum of the co-efficients are = 1 + 0 – 7 + 6 = 7 – 7 = 0

∴(x- 1) is a factor

Sum of co-efficients of even powers of x with constant = 0 + 6 = 6

Sum of coefficient of odd powers of x = 1 – 7 = -7

6 ≠ -7

∴ (x + 1) is not a factor

To find the other factors, let us use synthetic division.

∴ The factors are (x – 1), (x – 2), (x + 3)

∴ x^{3} + 0x^{2} – 7x + 6 = (x – 1) (x – 2) (x + 3)

(vi) x^{3} – 10x^{2} – x + 10

Let p (x) = x^{3} – 10x^{2} – x + 10

Sum of the co-efficients = 1 – 0 – 1 + 10

= 11 – 11 = 0

∴ (x – 1) is a factor

Sum of co-efficients of even powers of x with constant = -10 + 10 = 0

Sum of co-efficients of odd powers of = 1 – 1 = 0

∴(x + 1) is a factor

Synthetic division

∴ x^{3} + 10x^{2} – x + 10 = (x – 1) (x + 1) (x – 10)